Q. 409 - Intermediate Algebra

Question

In the following exercises, solve each system of equations. Use either substitution or elimination.

$\left\{\begin{array}{l}x+4y=6\\ -2x+y=-3\end{array}\right.$

Step-by-Step Solution

Answer

The solution for the system of the equation $\left\{\begin{array}{l}x+4y=6\\ -2x+y=-3\end{array}\right.$ is $(2,1)$

Step 1. Given

The system of equations $\left\{\begin{array}{l}x+4y=6\\ -2x+y=-3\end{array}\right.$

To find the solution of the system of equations.

Step 2. Solve the equations

Consider the system of equations:

   $x+4y=6$

$-2x+y=-3$

Solve the first equation for $x$,

           $x=6-4y$

Step 3. Solve for y

Substitute $x=6-4y$ in the second equation,

$-2(6-4y)+y=-3$

   $-12+8y+y=-3$

                    $9y=9$

Divide both sides of the equation by $9$,

                      $y=1$

Step 4. Solve for x

Substitute $y=1$ in the first equation,

  $x+4y=6$

$x+4\left(1\right)=6$

          $x=6-4$

          $x=2$

The solution of the equation is $(2,1)$

Step 5. Check the solution

Substitute $(2,1)$ in both the equations,

  $x+4y=6$

$2+4\left(1\right)=6$

           $6=6$ is true.

Then $-2x+y=-3$

       $-2\left(2\right)+1=-3$

                 $-3=-3$ is true.

Hence the solution has been checked.